Is the Given proof to the following Proof correct ? if not please do specify any mistakes.
$\\\textbf{Theorem.}$ Given that $R$ is a partial order on $A$, $B_1\subseteq A$, $B_2\subseteq A$, $x_1$ is the least upper bound of $B_1$ and $x_2$ is the least upper bound of $B_2$. If $B_1\subseteq B_2$ then $x_1Rx_2$
$\\\textbf{Proof.}$ Assume that $R$ is a partial order on $A$, $B_1\subseteq A$, $B_2\subseteq A$, $x_1$ is the least upper bound of $B_1$, $x_2$ is the least upper bound of $B_2$ and $B_1\subseteq B_2$. Since $x_1$ is the least upper bound of the set $B_1$ it follows that it is the $R-$smallest element of the set $U_1$ where $U_1 = \{a\in A|\forall s\in B_1(sRa)\}$ and considering that $B_1\subseteq B_2$ it follows that $\forall s\in B_1(sRx_2)$ but this implies that $x_2\in U_1$ and since $\forall q\in U_1(x_1Rq)$ then in particular we have $x_1Rx_2$. $\square$
Since x2 is an upper bound of B2 and B1 subset B2, x2 is an
upper bound of B1. As x1 is the least upper bound of B1, x1 <= x2.
Exercise. If y1, y2 are respectively, the greatest
lower bound of B1, B2, show y2 <= y1.